Simply clever. Congrats

What about sorted?

I like the different thikning

Haha love your solution ;)

As bkaes said, you don't want to surprise the user by modifying the input list. Lists and dictionaries are examples of data types that are mutable, so when you change an input list in the local function space, you are also changing the input list in the caller space. This can be a useful feature of these data types, but requires special care sometimes.

Additionally, you could have just returned [lst[0], lst[-1]] instead of creating the tempor variable.

Since min_max isn't documented, you expect the argument lst to hold the following property:

old_list = list(lst)
min_max(lst)
old_list == lst # element wise equal with same order

After all, min_max extracts information from lst, so it's not necessary to modify the list. Changing the list would surprise the user.

That being said, your solution runs in O(n log n) compared to the optimal O(n). If you don't know the O-notation: if the list has n elements, your solution needs about s * (n * log n) operations to get the minimum (where s is some constant), whereas an optimal solution only uses s' * n operations (where s' is another constant). If you're interested in O-notation, search for "Big O notation" or "Landau notation".

And why is it dangerous? Do you think I could lost some part of imput or is it a question of Best Practices?
I'm not really experienced programmer so thank you in advance for your answer.

A bit dangerous because you have modified the input list using the .sort() method. Sorting a copy or creating a new list using sorted avoids this situation.